Answer
$$3\ln \frac{4}{3}$$
Work Step by Step
Given $$\int_{8}^{27} \frac{d x}{x+x^{2 / 3}} $$
Let $$u^3=x\ \ \ \ \ \ 3 u^2du =dx $$
At $ x=8\ \ \to \ u=2$ and at $ x=27\ \ \to \ u=3$, so we have
\begin{align*}
\int_{8}^{27} \frac{d x}{x+x^{2 / 3}}&=\int_{2}^{3} \frac{3 u^2du}{u^3+u^{2 }}\\
&= \int_{2}^{ 3} \frac{3 du}{u +1}\\
&= 3\ln|u+1|\bigg|_{2}^{ 3}\\
&= 3\ln| 3+1|-3\ln |2+1|\\
&=3\ln \frac{4}{3}
\end{align*}
